How is capacitance related to the distance between the plates of a capacitor?

Assume that a pendulum used to drive a grandfather clock has a length L0=1.00m and a mass M at temperature T=20.00°C. It can be

Sedaia [141]

Answer:

The period will change a 0,036 % relative to its initial state

Explanation:

When the rod expands by heat its moment of inertia increases, but since there was no applied rotational force to the pendulum , the angular momentum remains constant. In other words:

ζ= Δ(Iω)/Δt, where ζ is the applied torque, I is moment of inertia, ω is angular velocity and t is time.

since there was no torque ( no rotational force applied)

ζ=0 → Δ(Iω)=0 → I₂ω₂ -I₁ω₁ = 0 → I₁ω₁ = I₂ω₂

thus

I₂/I₁ =ω₁/ω₂ , (2) represents final state and (1) initial state

we know also that ω=2π/T , where T is the period of the pendulum

I₂/I₁ =ω₁/ω₂ = (2π/T₁)/(2π/T₂)= T₂/T₁

Therefore to calculate the change in the period we have to calculate the moments of inertia. Looking at tables, can be found that the moment of inertia of a rod that rotates around an end is

I = 1/3 ML²

Therefore since the mass M is the same before and after the expansion

I₁ = 1/3 ML₁² , I₂ = 1/3 ML₂² → I₂/I₁ = (1/3 ML₂²)/(1/3 ML₁²)= L₂²/L₁²= (L₂/L₁)²

since

L₂= L₁ (1+αΔT) , L₂/L₁=1+αΔT , where ΔT is the change in temperature

now putting all together

T₂/T₁=I₂/I₁=(L₂/L₁)² = (1+αΔT) ²

finally

%change in period =(T₂-T₁)/T₁ = T₂/T₁ - 1 = (1+αΔT) ² -1

%change in period =(1+αΔT) ² -1 =[ 1+18×10⁻⁶ °C⁻¹ *10 °C]² -1 = 3,6 ×10⁻⁴ = 3,6 ×10⁻² % = 0,036 %

4 0

3 years ago

Find the mass of a sample of water if its temperature dropped 24.8 °C when it lost 870 J of

rosijanka [135]

Answer:

8.35 × $10^{-3}$ kg

Explanation:

Remember the formula for heat:

$Q=mc\Delta T$

You just need to solve for m, doing so you're left with:

$m=\dfrac{Q}{c\Delta T}$

Replace the variables with numbers and that's it.

7 0

2 years ago

An object, when pushed with a net force F, has an acceleration of 4 m/s . Now twice the force is applied to an object that has f

hjlf

Answer:

B. 2 m/s

B. Acceleration = 4.05 m/s² and Tension = 297.5 N.

Explanation:

A force is applied on a mass m whose acceleration is 4 m/s

Force = mass × acceleration

a = F/m = 4 m/s

4 m/s = F/m

F = 4 m/s (m)

If Force of 2F is applied on a mass of 4m ; it acceleration is as follows:

2F/4 m = F/ 2m

4m/s (m) / 2m = 2 m/s

a = 2 m/s

2.

Given that

mass $m_1$ = 30 kg

mass $m_2$ = 50 kg

$\mu$ = 0.1

From the question; we can arrive at two cases;

That :

$m_{2} a _ \ {net} }= m_2g - T$ ----- equation (1)

$m_{1} a _ \ {net} }= T - mg sin \theta - F$ ---- equation (2)

50 a = 50 g - T

30 a = T - 30 g sin 30 - 4 × 30 g cos 30

By summation

80 a = $[ 50 - 30 * \frac{1}{2} - 0.1 *30* \frac{\sqrt{3}}{2}]$ g

80 a = 32. 4 × 10 m/s ² (using g as 10m/s²)

80 a = 324 m/s ²

a = 324/80

a = 4.05 m/s²

From equation , replace a with 4.05

50 × 4.05 = 50 × 10 - T

T = 500 -202.5

T =297.5 N

8 0

3 years ago

T-Joe (65 kg) is running at 3 m/s. T-Brud (50 kg) is running at 4 m/s. What would be T-Joe's momentum?

Debora [2.8K]

Answer:

$P_J=195N$

Explanation:

From the question we are told that

$Mass\ of T-joe\ M_J=65\\Velocity\ of T-joe\ V_J=3m/s\\Mass of\ T-Brud\ M_B=50kg\\Velocity\ of T-Brud\ V_B=3m/s\\$

Generally the equation for momentum is mathematically given by

$P=mv$

Therefore

T-Joe momentum $P_J$

$P_J=65*3$

$P_J=195N$

5 0

2 years ago

Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating a

cupoosta [38]

Answer:

2.2 s

Explanation:

Using the equation for the period of a physical pendulum, T = 2π√(I/mgh) where I = moment of inertia of leg about perpendicular axis at one point = mL²/3 where m = mass of man = 67 kg and L = height of man = 1.83 m, g = acceleration due to gravity = 9.8 m/s² and h = distance of leg from center of gravity of man = L/2 (center of gravity of a cylinder)

So, T = 2π√(I/mgh)

T = 2π√(mL²/3 /mgL/2)

T = 2π√(2L/3g)

substituting the values of the variables into the equation, we have

T = 2π√(2L/3g)

T = 2π√(2 × 1.83 m/(3 × 9.8 m/s² ))

T = 2π√(3.66 m/(29.4 m/s² ))

T = 2π√(0.1245 s² ))

T = 2π(0.353 s)

T = 2.22 s

T ≅ 2.2 s

So, the period of the man's leg is 2.2 s

7 0

3 years ago